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Prove the inequality $sum x+6ge 2(sumsqrt{xy}) $

How prove this inequality $frac{a}{sqrt{b^2+b+1}}+frac{b}{sqrt{c^2+c+1}}+frac{c}{sqrt{a^2+a+1}}gesqrt{3}$How to prove $sqrt{frac{ab}{2a^2+bc+ca}}+sqrt{frac{bc}{2b^2+ca+ab}}+sqrt{frac{ca}{2c^2+ab+bc}}gefrac{81}{2}cdotfrac{abc}{(a+b+c)^3}$How prove this inequality $sumlimits_{cyc}frac{a^3}{b+c+d}ge dfrac{1}{3}$,if $sumlimits_{cyc}asqrt{bc}ge 1$Inequality $sqrt{1+x^2}+sqrt{1+y^2}+sqrt{1+z^2} le sqrt{2}(x+y+z)$Prove that $sumlimits_{cyc}sqrt[3]{a^2+4bc}geqsqrt[3]{45(ab+ac+bc)}$Prove that $sumlimits_{cyc}(4x^6+5x^5y)geqfrac{left(sumlimits_{cyc}(x^2+xy)right)^3}{8}$Inequality $frac{a}{sqrt{bc}}cdotfrac1{a+1}+frac{b}{sqrt{ca}}cdotfrac1{b+1}+frac{c}{sqrt{ab}}cdotfrac1{c+1}leqslantsqrt2.$Triangles geometric inequalitiesProve the inequality $sqrt{frac{a^2+1}{b+c}}+sqrt{frac{b^2+1}{a+c}}+sqrt{frac{c^2+1}{a+b}}ge 3$Prove $sum sqrt{frac{a^2}{6a^2+5ab+b^2}}le frac{sqrt{3}}{2}$

2

$begingroup$

Let $x;y;zin R^+$ such that $x+y+z+2=xyz$. Prove that $$x+y+z+6ge 2(sqrt{xy}+sqrt{yz}+sqrt{xz}) $$

---

This inequality is not homogeneous and look at the condition i thought that
i would substitute the variables $x;y;z$ such as:

+)If i need to solve $x^2+y^2+z^2+2xyz=1$, i will let $x=frac{2a}{sqrt{left(a+bright)left(a+cright)}}$

+)If i need to solve $xy+yz+xz+xyz=4$,i will let $x=frac{2a}{b+c}$

but failed. Please explain for me how can i get this substitution (if have a solution by substitution)

I also tried to solve it by $u,v,w$.Let $sum_{cyc} x=3u;sum_{cyc} xy;Pi_{cyc}a=w^3(3u+2=w^3;u,v,w>0 )$ then $ule w^3-3u$ or $4ule w^3$ but stuck (I am really bad at $uvw$)

algebra-precalculus inequality substitution a.m.-g.m.-inequality

share|cite|improve this question

edited Mar 15 at 13:39

Michael Rozenberg

109k1895200

asked Mar 14 at 12:20

nDLynknDLynk

23410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry just a typo
  $endgroup$
  – nDLynk
  Mar 14 at 12:39
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $x;y;zin R^+$ such that $x+y+z+2=xyz$. Prove that $$x+y+z+6ge 2(sqrt{xy}+sqrt{yz}+sqrt{xz}) $$

---

This inequality is not homogeneous and look at the condition i thought that
i would substitute the variables $x;y;z$ such as:

+)If i need to solve $x^2+y^2+z^2+2xyz=1$, i will let $x=frac{2a}{sqrt{left(a+bright)left(a+cright)}}$

+)If i need to solve $xy+yz+xz+xyz=4$,i will let $x=frac{2a}{b+c}$

but failed. Please explain for me how can i get this substitution (if have a solution by substitution)

I also tried to solve it by $u,v,w$.Let $sum_{cyc} x=3u;sum_{cyc} xy;Pi_{cyc}a=w^3(3u+2=w^3;u,v,w>0 )$ then $ule w^3-3u$ or $4ule w^3$ but stuck (I am really bad at $uvw$)

algebra-precalculus inequality substitution a.m.-g.m.-inequality

share|cite|improve this question

edited Mar 15 at 13:39

Michael Rozenberg

109k1895200

asked Mar 14 at 12:20

nDLynknDLynk

23410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry just a typo
  $endgroup$
  – nDLynk
  Mar 14 at 12:39
  

  
  
  
  

add a comment | 

$begingroup$

Let $x;y;zin R^+$ such that $x+y+z+2=xyz$. Prove that $$x+y+z+6ge 2(sqrt{xy}+sqrt{yz}+sqrt{xz}) $$

---

This inequality is not homogeneous and look at the condition i thought that
i would substitute the variables $x;y;z$ such as:

+)If i need to solve $x^2+y^2+z^2+2xyz=1$, i will let $x=frac{2a}{sqrt{left(a+bright)left(a+cright)}}$

+)If i need to solve $xy+yz+xz+xyz=4$,i will let $x=frac{2a}{b+c}$

but failed. Please explain for me how can i get this substitution (if have a solution by substitution)

I also tried to solve it by $u,v,w$.Let $sum_{cyc} x=3u;sum_{cyc} xy;Pi_{cyc}a=w^3(3u+2=w^3;u,v,w>0 )$ then $ule w^3-3u$ or $4ule w^3$ but stuck (I am really bad at $uvw$)

algebra-precalculus inequality substitution a.m.-g.m.-inequality

share|cite|improve this question

edited Mar 15 at 13:39

Michael Rozenberg

109k1895200

asked Mar 14 at 12:20

nDLynknDLynk

23410

$endgroup$

share|cite|improve this question

edited Mar 15 at 13:39

Michael Rozenberg

109k1895200

asked Mar 14 at 12:20

nDLynknDLynk

23410

share|cite|improve this question

edited Mar 15 at 13:39

Michael Rozenberg

109k1895200

asked Mar 14 at 12:20

nDLynknDLynk

23410

edited Mar 15 at 13:39

Michael Rozenberg

109k1895200

edited Mar 15 at 13:39

Michael Rozenberg

109k1895200

asked Mar 14 at 12:20

nDLynknDLynk

23410

asked Mar 14 at 12:20

nDLynknDLynk

23410

1 Answer
1

active

oldest

votes

2

$begingroup$

The condition gives
$$sum_{cyc}frac{1}{x+1}=1.$$
Now, let $x=frac{b+c}{a}$ and $y=frac{a+c}{b},$ where $a$, $b$ and $c$ are positives.

Thus, $z=frac{a+b}{c}$ and we need to prove that:
$$sum_{cyc}frac{b+c}{a}+6geq2sum_{cyc}sqrt{frac{(b+c)(a+c)}{ab}}$$ or
$$sum_{cyc}(a^2b+a^2c+2abc)geq2sum_{cyc}asqrt{bc(a+b)(a+c)},$$
which is true by AM-GM.

Indeed,
$$2sum_{cyc}asqrt{bc(a+b)(a+c)}=2sum_{cyc}asqrt{(ac+bc)(ab+bc)}leq$$
$$leqsum_{cyc}a(ac+bc+ab+bc)=sum_{cyc}(a^2b+a^2c+2abc).$$
Done!

share|cite|improve this answer

answered Mar 14 at 16:44

Michael RozenbergMichael Rozenberg

109k1895200

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  To be more to the point in the crucial substitution step, one can say ''let $a=frac{1}{x+1}$ and the analogues''; then $a,b,c>0, a+b+c=1$ and it will turn out that $x=frac{1}{a}-1=frac{b+c}{a}$ etc.
  $endgroup$
  – ΑΘΩ
  Mar 14 at 17:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, of course, it's the same.
  $endgroup$
  – Michael Rozenberg
  Mar 14 at 17:31
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  At the risk of being pedantic, I would like to respectfully disagree. Phrasing things like ''let $x=frac{b+c}{a}$ etc immediately raises the question ''do such $a, b, c$ affording the desired expressions actually exist?''. Of course this question can be tacitly answered in the background but a presentation such as the one suggested above clarifies everything from the onset and precludes any worries about such questions. Nevertheless, with all my fussing over matters, a very elegant solution!
  $endgroup$
  – ΑΘΩ
  Mar 14 at 17:36
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ΑΘΩ Nice question! Since $z(xy-1)=x+y+2>0,$ we obtain $xy>1$ and we can take $b=frac{(x+1)a}{y+1}$ and $c=frac{(xy-1)a}{y+1}.$
  $endgroup$
  – Michael Rozenberg
  Mar 14 at 17:47
  

  
  
  
  

add a comment | 

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2

$begingroup$

The condition gives
$$sum_{cyc}frac{1}{x+1}=1.$$
Now, let $x=frac{b+c}{a}$ and $y=frac{a+c}{b},$ where $a$, $b$ and $c$ are positives.

Thus, $z=frac{a+b}{c}$ and we need to prove that:
$$sum_{cyc}frac{b+c}{a}+6geq2sum_{cyc}sqrt{frac{(b+c)(a+c)}{ab}}$$ or
$$sum_{cyc}(a^2b+a^2c+2abc)geq2sum_{cyc}asqrt{bc(a+b)(a+c)},$$
which is true by AM-GM.

Indeed,
$$2sum_{cyc}asqrt{bc(a+b)(a+c)}=2sum_{cyc}asqrt{(ac+bc)(ab+bc)}leq$$
$$leqsum_{cyc}a(ac+bc+ab+bc)=sum_{cyc}(a^2b+a^2c+2abc).$$
Done!

share|cite|improve this answer

answered Mar 14 at 16:44

Michael RozenbergMichael Rozenberg

109k1895200

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  To be more to the point in the crucial substitution step, one can say ''let $a=frac{1}{x+1}$ and the analogues''; then $a,b,c>0, a+b+c=1$ and it will turn out that $x=frac{1}{a}-1=frac{b+c}{a}$ etc.
  $endgroup$
  – ΑΘΩ
  Mar 14 at 17:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, of course, it's the same.
  $endgroup$
  – Michael Rozenberg
  Mar 14 at 17:31
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  At the risk of being pedantic, I would like to respectfully disagree. Phrasing things like ''let $x=frac{b+c}{a}$ etc immediately raises the question ''do such $a, b, c$ affording the desired expressions actually exist?''. Of course this question can be tacitly answered in the background but a presentation such as the one suggested above clarifies everything from the onset and precludes any worries about such questions. Nevertheless, with all my fussing over matters, a very elegant solution!
  $endgroup$
  – ΑΘΩ
  Mar 14 at 17:36
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ΑΘΩ Nice question! Since $z(xy-1)=x+y+2>0,$ we obtain $xy>1$ and we can take $b=frac{(x+1)a}{y+1}$ and $c=frac{(xy-1)a}{y+1}.$
  $endgroup$
  – Michael Rozenberg
  Mar 14 at 17:47
  

  
  
  
  

add a comment | 

2

$begingroup$

The condition gives
$$sum_{cyc}frac{1}{x+1}=1.$$
Now, let $x=frac{b+c}{a}$ and $y=frac{a+c}{b},$ where $a$, $b$ and $c$ are positives.

Thus, $z=frac{a+b}{c}$ and we need to prove that:
$$sum_{cyc}frac{b+c}{a}+6geq2sum_{cyc}sqrt{frac{(b+c)(a+c)}{ab}}$$ or
$$sum_{cyc}(a^2b+a^2c+2abc)geq2sum_{cyc}asqrt{bc(a+b)(a+c)},$$
which is true by AM-GM.

Indeed,
$$2sum_{cyc}asqrt{bc(a+b)(a+c)}=2sum_{cyc}asqrt{(ac+bc)(ab+bc)}leq$$
$$leqsum_{cyc}a(ac+bc+ab+bc)=sum_{cyc}(a^2b+a^2c+2abc).$$
Done!

share|cite|improve this answer

answered Mar 14 at 16:44

Michael RozenbergMichael Rozenberg

109k1895200

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  To be more to the point in the crucial substitution step, one can say ''let $a=frac{1}{x+1}$ and the analogues''; then $a,b,c>0, a+b+c=1$ and it will turn out that $x=frac{1}{a}-1=frac{b+c}{a}$ etc.
  $endgroup$
  – ΑΘΩ
  Mar 14 at 17:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, of course, it's the same.
  $endgroup$
  – Michael Rozenberg
  Mar 14 at 17:31
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  At the risk of being pedantic, I would like to respectfully disagree. Phrasing things like ''let $x=frac{b+c}{a}$ etc immediately raises the question ''do such $a, b, c$ affording the desired expressions actually exist?''. Of course this question can be tacitly answered in the background but a presentation such as the one suggested above clarifies everything from the onset and precludes any worries about such questions. Nevertheless, with all my fussing over matters, a very elegant solution!
  $endgroup$
  – ΑΘΩ
  Mar 14 at 17:36
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ΑΘΩ Nice question! Since $z(xy-1)=x+y+2>0,$ we obtain $xy>1$ and we can take $b=frac{(x+1)a}{y+1}$ and $c=frac{(xy-1)a}{y+1}.$
  $endgroup$
  – Michael Rozenberg
  Mar 14 at 17:47
  

  
  
  
  

add a comment | 

$begingroup$

The condition gives
$$sum_{cyc}frac{1}{x+1}=1.$$
Now, let $x=frac{b+c}{a}$ and $y=frac{a+c}{b},$ where $a$, $b$ and $c$ are positives.

Thus, $z=frac{a+b}{c}$ and we need to prove that:
$$sum_{cyc}frac{b+c}{a}+6geq2sum_{cyc}sqrt{frac{(b+c)(a+c)}{ab}}$$ or
$$sum_{cyc}(a^2b+a^2c+2abc)geq2sum_{cyc}asqrt{bc(a+b)(a+c)},$$
which is true by AM-GM.

Indeed,
$$2sum_{cyc}asqrt{bc(a+b)(a+c)}=2sum_{cyc}asqrt{(ac+bc)(ab+bc)}leq$$
$$leqsum_{cyc}a(ac+bc+ab+bc)=sum_{cyc}(a^2b+a^2c+2abc).$$
Done!

share|cite|improve this answer

answered Mar 14 at 16:44

Michael RozenbergMichael Rozenberg

109k1895200

$endgroup$

share|cite|improve this answer

answered Mar 14 at 16:44

Michael RozenbergMichael Rozenberg

109k1895200

answered Mar 14 at 16:44

Michael RozenbergMichael Rozenberg

109k1895200

answered Mar 14 at 16:44

Michael RozenbergMichael Rozenberg

109k1895200